Proof

Write H = G_x. We split the proof in the following steps.

The map t is well defined.

The map t is injective.

The map t is surjective.

For each h G, we have t · L_h = h · t .

If G is finite, then |G/H| = |X|.

Suppose g and g' are in the same coset gH. Then there is h H, such that g' = gh. As H = G_x, we have h(x) = x, whence g'(x) = gh(x) = g(x). This proves that the assigment t(gH) = g(x) does not depend on the choice of g' gH.

Suppose g, g' G satisfy t(gH) = t(g'H). Then g(x) = g'(x), so x = g^-1g'(x), that is, g^-1g' G_x = H. This shows g^-1g'H = H, and so g'H = gH.

Hence t is injective.

Let y X. As f is transitive, there is g G with y = g(x). But then t(gH) = y.

Hence t is surjective.

Let g G. Then t · L_h(gH) = t (hgH) = hg(x) = h(t(gH)) = h · t(gH).

Hence, t · L_h = h · t.